Cases in which the limiting value of a function $f(x)$ (as $x\to c$) is not equal to $f(c)$? Can anyone state some cases such as mentioned in the title. I have tried to look for them on google and my textbook, but cant find any examples. And incidentally how can this fact be ever true? does it have something to with the function not  being defined at the limiting point $c$?
 A: Consider this example
$$
\begin{align}
f(x) = \begin{cases} 
1 & \text{ if } x = 0 \\
 0 & \text{ if } x\neq 0
\end{cases}
\end{align}
$$
Here
$$
f(0) = 1
$$
but
$$
\lim_{x\to 0} f(x) = 0.
$$
I think your confusion comes from the fact that most functions you know (and can visualize) are continuous functions. Being continuous basically means that the graph is connected, that is, you can draw the graph without lifting the pen. For such functions you will always have that $\lim_{x\to c} f(x) = f(c)$. So to give an example of a function that does not satisfy this formula, you have to think about a non-continuous (or discontinuous function). The graph of the function above follows the $x$-axis except at the one point where $x=0$. Here is it is equal to $1$. So there is a jump right at that point. The value of the function anywhere close to $x=0$ is $0$, and only at $x=0$ is the function not $0$. 
A: There are artificial examples created by piecewise definitions. (Richard Feynman once expressed surprise that anybody thought such things are functions.)  I'll try to give a more serious example: You are walking past a building with a conventional rectangular shape.  It is on your left.  The distance from you to the rightmost point on the building that you can see changes in a continuous manner until you pass the corner, and then it abruptly increases; it has a jump discontinuity.
A: $$
\begin{align}
f(x) = \begin{cases} 
x^2-2x & \text{ if } x \not= \frac12 \\
 3 & \text{ if } x= \frac12 
\end{cases}
\end{align}
$$

